I'm going crazy over this problem. I've spent so much time on it that it sux. I've included a picture for your pleasure. A small spherical ball of radius r = 2.2 cm rolls without slipping down a ramp and around a loop-the-loop of radius R = 2.7 m. The ball has mass M = 378 g. **************************** a) How high above the top of the loop must it be released in order that the ball just makes it around the loop? b) Repeat problem (a) for a disk. Find the ratio of the heights h for the two cases. ----------------------- Okay,I'm just stuck on part a. If I figure that out, rest should be peice of cake. I know that I'd have to use conservation of energy. The ball starts off with potential energy, and that converts into translational and rotational energy. So the equation that I've set up is: mgh=.5MV^2+.5*I*w^2 (equation 1) where I is moment of inertia and w is omega. Since it is a sphere, the I= 2/5Mr^2 So right now I know m=.378kg g=9.8m/s/s r=.022m but not omega or velocity, but I can calculate those two. At the top of the loop, there is a normal force and gravity force acting on the ball. So thats 2*mg, or 7.40N. Centripetal force = m(V^2/R) Solving that, I get velocity equals 7.27 m/s Then v=w*r Using that velocity number, I get omega=330.45/s So I plugged in all my values into equation 1, solve for H, and I get H=1.43m, but that isn't right. How do I solve this? I think I'm on the right track, but something along the way is screwing me over!